MQ
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KC
2
8 tháng 11 2016
\(A=\frac{x^{39}+x^{36}+x^{33}+...+x^3+1}{x^{40}+x^{38}+x^{36}+...+x^2+1}\)
Đặt \(C=x^{39}+x^{36}+x^{33}+...+x^3+1\)
\(x^3.C=x^{42}+x^{39}+x^{36}+...+x^3\)
\(\left(x^3-1\right)C=x^{42-1}\)
\(C=\frac{x^{42}-1}{x^3-1}\)
Đặt \(D=x^{40}+x^{38}+x^{36}+....+x^2+1\)
\(x^2.D=x^{42}+x^{40}+x^{38}+x^{36}+....+x^2\)
\(\left(x^2-1\right).D=x^{42}-1\)
\(D=\frac{x^{42}-1}{x^2-1}\)
Ta có :
\(C:D=\frac{x^{42}-1}{x^3-1}:\frac{x^{42}-1}{x^2-1}\)
\(C:D=\frac{x^2-1}{x^3-1}\)
\(C:D=\frac{x+1}{x^2+x+1}\)
Ta có : \(A=C:D=\frac{x+1}{x^2+x+1}\)
Vậy ...........
21 tháng 6 2017
Rút gọn.
\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)
ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)
\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)
=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)
Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
21 tháng 6 2017
Ta có:
\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)
\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)
\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)
\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)
tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
8 tháng 4 2017
Bài này cậu hỏi lâu rồi nên không biết cậu muốn biết lời giải bài đó nữa không vậy?
29 tháng 4 2020
\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)
\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)
\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)
<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)
<=> x=-40
Vậy x=-40
3 tháng 8 2019
\(40^2-39^2+38^2-37 ^2+...+2^2-1^2\)
= \(\left(40+39\right)\left(40-39\right)+\left(38+37\right)\left(38-37\right)+....+\left(2+1\right)\left(2-1\right)\)
= \(79.1+75.1+....+3.1\)
= \(79+75+....+3\)
= \(\left(79+3\right)\left[\left(79-3\right):4+1\right]:2\)
= \(82.20:2\)
= \(820\)
\(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
=> \(9x^2-6x+1+2x^2+12x+18-11x^2+11=6\)
=> \(6x+30=6\)
=> \(6x=6-30\)
=> \(6x=-24\)
=> \(x=-24:6=-4\)
3 tháng 8 2019
\(\text{a) }40^2-39^2+38^2-37^2+...+2^2-1^2\)
\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1.79+1.75+...+1.3\)
\(=79+75+...+3\)
\(\text{Từ 3 đến 79 có: (79 - 3) : 2 + 1 = 39 (số hạng)}\)
\(\text{Tổng là: }\frac{\left(79+3\right)\times39}{2}=1599\)
\(\text{b) }\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow\left(9x^2-6x+1\right)+2\left(x^2+6x+9\right)+11\left(1-x^2\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18+11-11x^2=6\)
\(\Leftrightarrow\left(9x^2+2x^2-11x^2\right)+\left(-6x+12x\right)+\left(1+18+11\right)=6\)
\(\Leftrightarrow6x+30=6\)
\(\Leftrightarrow6x=6-30\)
\(\Leftrightarrow6x=-24\)
\(\Leftrightarrow x=-4\)
18 tháng 8 2016
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
24 tháng 2 2020
a, Ta có : \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=-5+5=0\)
=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
=> \(424-x=0\)
=> \(x=424\)
Vậy phương trình có nghiệm là x = 424 .
b, Ta có : \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
=> \(\frac{x+1}{2014}+1+\frac{x+3}{2012}+1=\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}-\frac{x+2015}{2010}-\frac{x+2015}{2009}=0\)
=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> \(x+2015=0\)
=> \(x=-2015\)
Vậy phương trình có nghiệm là x = -2015 .
24 tháng 2 2020
a) \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
<=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{40}\right)=0\)
<=> 424 - x = 0
<=> x = 424
Vậy S = {424}
b) \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
<=> x + 2015 = 0
<=> x= -2015
Vậy S = {-2015}